Minimum Lines to Represent a Line Chart

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Arrays

You are given a 2D integer array stockPrices where stockPrices[i] = [day_i, price_i] indicates the price of the stock on day day_i is price_i. A line chart is created from the array by plotting the points on an XY plane with the X-axis representing the day and the Y-axis representing the price and connecting adjacent points. One such example is shown below:

Return the minimum number of lines needed to represent the line chart.

Example 1:

Input: stockPrices = [[1,7],[2,6],[3,5],[4,4],[5,4],[6,3],[7,2],[8,1]]
Output: 3
Explanation:
The diagram above represents the input, with the X-axis representing the day and Y-axis representing the price.
The following 3 lines can be drawn to represent the line chart:
- Line 1 (in red) from (1,7) to (4,4) passing through (1,7), (2,6), (3,5), and (4,4).
- Line 2 (in blue) from (4,4) to (5,4).
- Line 3 (in green) from (5,4) to (8,1) passing through (5,4), (6,3), (7,2), and (8,1).
It can be shown that it is not possible to represent the line chart using less than 3 lines.

Example 2:

Input: stockPrices = [[3,4],[1,2],[7,8],[2,3]]
Output: 1
Explanation:
As shown in the diagram above, the line chart can be represented with a single line.

Constraints:

1 <= stockPrices.length <= 10⁵
stockPrices[i].length == 2
1 <= day_i, price_i <= 10⁹
All day_i are distinct.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected data type of the input array's elements? Are they integers or floating-point numbers?
What are the possible ranges for the y-values in the input points? Can they be negative, zero, or very large?
If the input array has fewer than two points, what should the function return?
If all points lie on the same line, should I return 1 or handle it as a special case?
Is the input array guaranteed to be sorted by x-coordinate, or do I need to sort it first?

Brute Force Solution

Approach

The brute force approach for this problem is to try every possible combination of line segments to represent the chart. We will consider each possible way to split the data points into consecutive groups, where each group represents a line segment.

Here's how the algorithm would work step-by-step:

Start by assuming that the first two data points define the first line.
Then, for each remaining data point, check if it falls on the same line as the previous two. If so, keep it on the same line.
If a data point does NOT fall on the same line, then start a new line segment from the previous point to the current point.
Record the number of lines required for this specific way of connecting the dots.
Repeat this process, starting with the first three data points defining the first line, then the first four, and so on, each time checking if subsequent points can be added to the line, and counting the lines required.
Continue this process, considering every possible starting point and length for the first line.
After trying all the possibilities, choose the representation that uses the fewest number of lines.

Code Implementation

def minimum_lines_brute_force(points):
    number_of_points = len(points)
    if number_of_points <= 2:
        return 1

    minimum_lines = number_of_points - 1

    for first_line_end in range(1, number_of_points):
        number_of_lines = 1
        current_line_start = 0
        current_line_end = first_line_end

        while current_line_end < number_of_points - 1:
            next_point_index = current_line_end + 1
            
            # Check if the next point is on the current line.
            if not is_on_line(points[current_line_start], points[current_line_end], points[next_point_index]):

                # Start a new line if the next point doesn't fit.
                number_of_lines += 1
                current_line_start = current_line_end

            current_line_end += 1

        minimum_lines = min(minimum_lines, number_of_lines)

    return minimum_lines

def is_on_line(point1, point2, point3):
    if point1[0] == point2[0]:
        return point3[0] == point1[0]
    
    # Calculate the slope of the line.
    slope = (point2[1] - point1[1]) / (point2[0] - point1[0])
    
    # Calculate the y-intercept.
    y_intercept = point1[1] - slope * point1[0]

    # Check if point3 lies on the line.
    return abs(point3[1] - (slope * point3[0] + y_intercept)) < 1e-6

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible starting points for the first line segment, which is O(n). For each starting point, it considers all possible lengths for the initial line segment, also O(n). Then, for each of these combinations, it iterates through the remaining points to determine if they fall on the same line, taking O(n) in the worst case. Therefore, the overall time complexity is O(n * n * n) which is O(n^3).

Space Complexity

O(1) – The provided brute force approach iterates through possible line segment combinations and counts the lines required for each. It doesn't explicitly create any auxiliary data structures like arrays, lists, or hash maps to store intermediate results or visited states. The algorithm primarily uses a fixed number of variables to track the current line and count lines. Therefore, the auxiliary space used remains constant regardless of the number of data points, N, making the space complexity O(1).

Optimal Solution

Approach

To find the fewest lines to represent a line chart, we should greedily group points into lines. We check if consecutive points can form a straight line with the current line and if not, we start a new one.

Here's how the algorithm would work step-by-step:

Start with the first two points, assuming they form the first line.
Consider the next point and check if it lies on the same line as the previous two points.
If the new point does lie on the same line, continue to the next point and repeat the check.
If the new point does not lie on the same line, count the current line and start a new line with the last point that was on the previous line and the new point.
Repeat this process until all points have been considered.
The final count represents the minimum number of lines required.

Code Implementation

def minimum_lines(points):
    number_of_points = len(points)
    if number_of_points <= 1:
        return number_of_points

    number_of_lines = 1
    first_point_index = 0
    second_point_index = 1

    for current_point_index in range(2, number_of_points):
        # Check if the current point lies on the same line
        x1, y1 = points[first_point_index]
        x2, y2 = points[second_point_index]
        x3, y3 = points[current_point_index]

        if (y2 - y1) * (x3 - x2) != (y3 - y2) * (x2 - x1):
            # If not, increment line count and reset
            number_of_lines += 1

            first_point_index = current_point_index - 1

            second_point_index = current_point_index

    return number_of_lines

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the points of the line chart once. Inside the loop, it checks if a new point lies on the current line formed by the previous two points. This check involves a constant number of arithmetic operations (cross-product calculation). Therefore, the time complexity is directly proportional to the number of points, n. Since the number of operations is constant per point, the overall time complexity is O(n).

Space Complexity

O(1) – The provided plain English explanation describes an iterative process that primarily involves comparing consecutive points and updating a line count. It doesn't mention the creation of any auxiliary data structures like lists, hash maps, or arrays to store intermediate results. The algorithm only needs to keep track of a few variables, such as the current line count and potentially indices to keep track of the points forming the current line. Therefore, the space used is constant and independent of the number of points (N) in the input line chart.

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no lines are needed to represent an empty chart.
Input array with only one point	Return 0 since a single point doesn't form a line.
Input array with only two points	Return 1 since two points define a single line segment.
All points in the array have the same y-value	Return 1 since all points lie on a horizontal line.
Points are collinear but not sorted by x-value	The algorithm needs to sort to correctly determine collinearity by checking slopes between adjacent points after sorting.
Large input array causing potential integer overflow in slope calculation	Use long or double to store slope values to avoid integer overflow during calculations.
Adjacent points have the same x-coordinate (vertical line)	Handle the case of infinite slope by using a flag or a very large number as slope representation or by checking X values before slope calculation.
Floating-point precision issues when comparing slopes	Use a small tolerance value (epsilon) when comparing floating-point slopes to account for precision errors.

Empty input array

How to Handle:

Return 0 since no lines are needed to represent an empty chart.

Input array with only one point

How to Handle:

Return 0 since a single point doesn't form a line.

Input array with only two points

How to Handle:

Return 1 since two points define a single line segment.

All points in the array have the same y-value

How to Handle:

Return 1 since all points lie on a horizontal line.

Points are collinear but not sorted by x-value

How to Handle:

The algorithm needs to sort to correctly determine collinearity by checking slopes between adjacent points after sorting.

Large input array causing potential integer overflow in slope calculation

How to Handle:

Use long or double to store slope values to avoid integer overflow during calculations.

Adjacent points have the same x-coordinate (vertical line)

How to Handle:

Handle the case of infinite slope by using a flag or a very large number as slope representation or by checking X values before slope calculation.

Floating-point precision issues when comparing slopes

How to Handle:

Use a small tolerance value (epsilon) when comparing floating-point slopes to account for precision errors.